A276053 Triangle read by rows: T(n,k) is the number of compositions of n with parts in {1,2,4,6,8,10,...} and having asymmetry degree equal to k (n>=0; 0<=k<=floor(n/3)).
1, 1, 2, 1, 2, 4, 2, 2, 8, 7, 8, 4, 3, 26, 4, 13, 24, 24, 6, 66, 28, 8, 23, 62, 104, 8, 10, 158, 120, 64, 42, 148, 352, 80, 16, 19, 350, 416, 344, 16, 75, 334, 1052, 448, 160, 33, 756, 1252, 1440, 208, 32, 136, 726, 2860, 1936, 1024, 32, 61, 1578, 3448, 5176, 1440, 384, 244, 1534, 7312, 7056, 5072, 512, 64
Offset: 0
Examples
Row 4 is [4,2] because the compositions of 4 with parts in {1,2,4,6,8,...} are 4, 22, 211, 121, 112, and 1111, having asymmetry degrees 0, 0, 1, 0, 1, and 0, respectively. Triangle starts: 1; 1; 2; 1,2; 4,2; 2,8; 7,8,4.
References
- S. Heubach and T. Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
Links
- Krithnaswami Alladi and V. E. Hoggatt, Jr. Compositions with Ones and Twos, Fibonacci Quarterly, 13 (1975), 233-239.
- V. E. Hoggatt, Jr. and Marjorie Bicknell, Palindromic compositions, Fibonacci Quart., Vol. 13(4), 1975, pp. 350-356.
Programs
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Maple
G := (1-z^4)*(1+z-z^3)/(1-2*z^2-2*t*z^3-z^4+(3-2*t)*z^6+2*t*z^7-z^8): Gser := simplify(series(G, z = 0, 30)): for n from 0 to 25 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 25 do seq(coeff(P[n], t, j), j = 0 .. degree(P[n])) end do; # yields sequence in triangular form
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Mathematica
Table[TakeWhile[BinCounts[#, {0, 1 + Floor[n/3], 1}], # != 0 &] &@ Map[Total, Map[Map[Boole[# >= 1] &, BitXor[Take[# - 1, Ceiling[Length[#]/2]], Reverse@ Take[# - 1, -Ceiling[Length[#]/2]]]] &, Flatten[Map[Permutations, DeleteCases[IntegerPartitions@ n, {_, a_, _} /; Nor[a == 1, EvenQ@ a]]], 1]]], {n, 0, 18}] // Flatten (* Michael De Vlieger, Aug 28 2016 *)
Formula
G.f.: G(t,z) = (1-z^4)*(1+z-z^3)/(1-2*z^2-2*t*z^3-z^4+(3-2*t)*z^6+2*t*z^7-z^8). In the more general situation of compositions into a[1]=1} z^(a[j]), we have G(t,z) = (1 + F(z))/(1 - F(z^2) - t*(F(z)^2 - F(z^2))). In particular, for t=0 we obtain Theorem 1.2 of the Hoggatt et al. reference.
Comments